How do I convert tuple of tuples to list in one li

问题:

query = 'select mydata from mytable'
cursor.execute(query)
myoutput = cursor.fetchall() 
print myoutput

(('aa',), ('bb',), ('cc',))

Why is it (cursor.fetchall) returning a tuple of tuples instead of a tuple since my query is asking for only one column of data?

What is the best way of converting it to ['aa', 'bb', 'cc'] ?

I can do something like this :

mylist = []
myoutput = list(myoutput)
for each in myoutput:
   mylist.append(each[0])

I am sure this isn't the best way of doing it. Please enlighten me!

回答1:

This works as well:

>>> tu = (('aa',), ('bb',), ('cc',))
>>> import itertools
>>> list(itertools.chain(*tu))
['aa', 'bb', 'cc']

Edit Could you please comment on the cost tradeoff? (for loop and itertools)

Itertools is significantly faster:

>>> t = timeit.Timer(stmt="itertools.chain(*(('aa',), ('bb',), ('cc',)))")
>>> print t.timeit()
0.341422080994
>>> t = timeit.Timer(stmt="[a[0] for a in (('aa',), ('bb',), ('cc',))]")
>>> print t.timeit()
0.575773954391

Edit 2 Could you pl explain itertools.chain(*)

That * unpacks the sequence into positional arguments, in this case a nested tuple of tuples.

Example:

>>> def f(*args):
...    print "len args:",len(args)
...    for a in args:
...       print a
... 
>>> tu = (('aa',), ('bb',), ('cc',))
>>> f(tu)
len args: 1
(('aa',), ('bb',), ('cc',))
>>> f(*tu)
len args: 3
('aa',)
('bb',)
('cc',)

Another example:

>>> f('abcde')
len args: 1
abcde
>>> f(*'abcde')
len args: 5
a
b
c
d
e

See the documents on unpacking.

回答2:

You could do

>>> tup = (('aa',), ('bb',), ('cc',))
>>> lst = [a[0] for a in tup]
>>> lst
['aa', 'bb', 'cc']

回答3:

Why is cursor.fetchall() returning a tuple of tuples instead of a tuple since my query is asking for only one column of data?

The outer tuple is the complete result; each inner tuple represents one record in that result; because you asked for only one field, each inner tuple has only one element.

What is the best way of converting it to ['aa', 'bb', 'cc'] ?

There are several ways, and which is 'best' depends on what you are doing...

Simple list comprehension:

mylist = [each[0] for each in myoutput]

Simple generator (saves on memory usage):

mygen = (each[0] for each in myoutput)
for result in mygen:
    print result

If you just need to process the items in myoutput, you could also do

for each, in myoutput:
    print each

If you have profiled your code and discovered that this is a bottleneck, then you can go for less readable but faster:

import itertools
mylist = list(itertools.chain(*myoutput))

or, again if you just need to process it:

import itertools
for result in itertools.chain(*myoutput):
    print result

回答4:

What you are doing is correct but more concise and may be better performing could be

>>> [item for item, in (('aa',), ('bb',), ('cc',)) ]
['aa', 'bb', 'cc']

or if you hate for keyword, you can use map

>>> map(lambda a:a[0], (('aa',), ('bb',), ('cc',)) )
['aa', 'bb', 'cc']

and here is another way

>>> reduce(lambda a, b:a+b, (('aa',), ('bb',), ('cc',)) )
('aa', 'bb', 'cc')

though IMO list comprehension is most readable

回答5:

This works:

>>> tups=(('aa',), ('bb',), ('cc',))
>>> list(*(zip(*tups)))
['aa', 'bb', 'cc']

Explanation:

1) *tups unpacks the nested tuples  ('aa'),('bb'),('cc')
2) zip produces a list of a single tuple with all the elements: [('aa','bb','cc')]
3) * unpacks that  into  'aa', 'bb', 'cc'
4) creates a list from that unpacking.

You could also do:

>>> list(zip(*tups)[0])
['aa', 'bb', 'cc']

回答6:

Do a list comprehension like this:

mylist = [ x[0] for x in myoutput ]